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Saturday, 31 October

JEE Main 2022 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 30 June 2022, Shift 1, Physics Q44

Question 44 of 90 in this shift, Physics question 14 of 30, Section A.

An expression for oscillating electric field in a plane electromagnetic wave is given as Ez\mathrm{E_z}=300 sin(5π×103x−3π×1011t)(5\pi\times10^3 x-3\pi\times10^{11}\mathrm{t}) Vm−1\mathrm{Vm^{-1}} Then, the value of magnetic field amplitude will be : (Given : speed of light in Vacuum c=3×108 ms−13\times10^8\ \mathrm{ms^{-1}})
  1. (1)1×10−61\times10^{-6} T
  2. (2)5×10−65\times10^{-6} TOfficial answer
  3. (3)18×10918\times10^{9} T
  4. (4)21×10921\times10^{9} T

Official answer

Option 2

NTA final key.

Chapter
Electromagnetic Waves
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.